Hydrocarbon Processing - April 2021 - GP-19

SPECIAL FOCUS: GREEN TECHNOLOGIES
The emission rate, E, of pollutants like NOX from a combustion system must first be defined. A combustion system is used
to generate certain power or heat, thereby producing pollutants.
The higher the power is generated by the system, the more pollutants are produced. Since all pollutants are discharged to air
through the tail flame, the total mass flowrates of the pollutants
can be calculated from the volume mass concentration of pollutants, mg/m3, and the volume flowrate of the tail flame.
E can be defined for a certain pollutant as the total mass
flowrate for this pollutant in the tail flame, divided by the
power generated by the combustion system. This E is listed in
mg/kWh. In this article, E is used for emission of NOX , but it
also can be used for emission of other pollutants, such as CO,
SO2 , SO3 , etc.
Now the volume flowrate of the tail flame of a combustion
system can be calculated. For the sake of simplicity, it is assumed that the concentrations of CO, NO, NO2 , SO2 , etc. in
the combustion products are negligible compared to the concentrations of O2 , CO2 , H2O and N2. At the same time, it is assumed that there is no water vapor condensation, at least at the
measurement point in the tail flame generated by the combustion system. Based on the authors' many years of experience
in the research and development of various types of heaters in
the U.S. and China, these two assumptions should be valid for
general combustion systems, such as infrared heaters, unitary
heaters, home heaters, etc.
In air, the composition of O2 is 20.95%, while N2 , Ar and
other elements make up 79.05% in volume.16 If the volume
flowrate of O2 is 1 for a combustion system, then the ratio of
the volume flowrate of N2 , Ar, etc. to the volume flowrate of
O2 is 79.05 ÷ 20.95 = 3.77. Since gas components like Ar do
not participate in the reaction, for the sake of brevity, only N2 is
used to represent these components.
Consider the combustion of methane (Eq. 10):

Va + 2Va + 7.54Va + y(Va + 3.77Va )(13)
Therefore, the total flow of the tail flame produced by the combustion system will be (10.54 + 4.77y)Va.
The y in Eqs. 12 and 13 can be obtained from the concentration of CO2 measured in the tail flame or the concentration
of O2 measured. First, the measured CO2 concentration is measured, as shown in Eq. 14:
Va
1
CO2measured =
=
(14)
(10.54 + 4.77 y ) × Va 10.54 + 4.77 y
The result is applied to Eq. 15:
1

− 10.54
CO2measured
(15)
y =
4.77
In this way, the volumetric flow of the tail flame can be represented as shown in Eq. 16:

Volume flowrate of the tail flame = 10.54Va +
4.77yVa = (1 / CO2 measured )Va

The power of the combustion system is generated by the consumption of CH4 at a specified flowrate (Va) and heat of combustion (HV), with the power of combustion = HV × Va . Since
the mass emission rate of NOX is X, Eq. 17 is used to calculate E:
E = (Volume mass concentration of emission
of NOX × Volume flow of tail flame) /
Power of combustion system

CH4 + 2(O2 + 3.77 N2 ) = CO2 + 2H2O + 7.54 N2(11)
If Va represents the volume rate of CH4 consumption, then
when the reaction product is cooled to normal temperature (assuming that the water vapor is not condensed at this time), the
volume rate of the tail flame generated will be Va + 2Va + 7.54Va
= 10.54Va. According to combustion theory,18 to avoid significant production of CO and carbon, there should be a certain
excess of O2 in combustion systems. Considering this, the reaction formula in Eq. 11 can be written as shown in Eq. 12:
CH4 + (2 + y)(O2 + 3.77N2 ) = CO2 + 2H2O +
7.54N2 + y(O2 + 3.77N2 )

(12)

It is assumed that y(O2 + 3.77N2) is O2 and corresponding
N2 in the excess air. If Va is the rate at which the volume of fuel
gas CH4 is consumed, then the volume rate at which the reactants disappear and the volume rate of the products generated
in the tail flame (when cooled to room temperature) can be calculated as shown in Eq. 13:
Va + 2(Va + 3.77Va ) + y(Va + 3.77Va ) and

(17)

Substituting Eq. 16 and the power of combustion (HV × Va)
into the formula results in Eq. 18:
E = X / (CO2measured × HV)

(18)

Using the same procedure, from the measured O2 concentration (O2measured ), E can be found, as shown in Eq. 19:

CH4 + 2O2 = CO2 + 2H2O(10)
If the reaction or combustion occurs in air, then Eq. 11 can
be used:

(16)



E =

(1

10.54 X
4.77O2measured )

HV

(19)

In Eq. 19, HV can be the low-combustion value (LHV) of CH4
or the high combustion value (HHV) of CH4, which is determined by the condensation of water vapor in the tail flame. The
authors suggest using the LHV because it is assumed that the
measurement is conducted before the water condensation,19
resulting in Eq. 20:
106 J
m
(20)
m3
3,600 sec
= 9.97 kWh/m 3
3,600 sec
LHV = 35.9

106 J

3

(

= 35.9

)

It is known that 1 hr = 3,600 sec, and 1 joule/sec = 1 W. If
pressure is assumed at 1 atm and temperature is assumed at
20°C (293K), and if X = 1.25 qx mg/m3 is used, then Eqs. 21
and 22 can be calculated:
E = (0.125x)q ÷ CO2measured mg/kWh

(21)

1.32xq
E =
mg/kWh
(22)
1 − 4.77O2measured
Gas Processing & LNG | MARCH/APRIL 2021 19
Hydrocarbon Processing - April 2021

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